# Monotone likelihood ratio

• May 8th 2011, 09:08 PM
Monotone likelihood ratio
I am trying to show that the monotone likelihood ratio for a uniform distribution [0,theta] is the maximum value (X1,...Xn). I know how to compute the monotone likelihood ratio by using fn(x given theta1)/fn(x given theta2), but then how do i show this is the maximum (X1,...Xn)?? Thanks
• May 8th 2011, 09:30 PM
matheagle
do you know that the likelihood function is

$\displaystyle {1\over \theta^n} I_{X_{(1)}>0} I_{X_{(n)}<\theta}$
• May 8th 2011, 09:33 PM
Yes so the MLR becomes (theta2^n)/(theta1^n). But what is the step that shows that that is the maximum (X1,...Xn)??
• May 8th 2011, 09:37 PM
matheagle
I'm not quite sure what your two theta's are
are you testing theta1 vs theta2?
• May 8th 2011, 09:42 PM
Monotone likelihood ratio - Wikipedia, the free encyclopedia

It is like the formula on this page only our class uses theta1 as f(x) and theta2 as g(x)
• May 8th 2011, 10:29 PM
Sambit
Suppose $\displaystyle \theta_2>\theta_1$. Then the ratio becomes $\displaystyle \frac{\theta_1^n}{\theta_2^n}\frac{I(X_{(n)},\thet a_2)}{I(X_{(n)},\theta_1)}$ [where $\displaystyle I(a,b)=1 ,$ if $\displaystyle a\leq b$ and 0 otherwise.]

$\displaystyle =\frac{\theta_1^n}{\theta_2^n}\frac{1}{1}$ , if $\displaystyle 0<X_{(n)}<\theta_1$
and $\displaystyle \frac{\theta_1^n}{\theta_2^n}\frac{1}{0}$ if $\displaystyle \theta_1<X_{(n)}<\infty$

that is $\displaystyle = \frac{\theta_1^n}{\theta_2^n}$ ,if $\displaystyle 0<X_{(n)}<\theta_1$
and $\displaystyle = \infty$ if $\displaystyle \theta_1<X_{(n)}<\infty$

So as $\displaystyle X_{(n)}$ increases, the ratio also increases. Hence MLR in $\displaystyle X_{(n)}$.
• May 8th 2011, 10:46 PM
Thank you, but how do i show the statistic is the maximum(X1,...Xn)
• May 9th 2011, 04:45 AM
Sambit
Quote:

How do you write the pdf (or the joint pdf) of a $\displaystyle U(0,\theta)$ distribution?
You are missing something. The joint pdf is given by $\displaystyle f(x_1,x_2,....,x_n)=\frac{1}{\theta^n},$ if $\displaystyle 0<X_i<\theta$ $\displaystyle \forall i=1(1)n$. Since all $\displaystyle x_i$ are less than $\displaystyle \theta$, it is enough to ensure that max (X1,X2,....,Xn), which is denoted by $\displaystyle X_{(n)}$, is always less than $\displaystyle \theta$. Hence the joint pdf becomes $\displaystyle f(x_1,x_2,....,x_n)=\frac{1}{\theta^n},$ if $\displaystyle 0<X_{(n)}<\theta$. Now you move to the ratio.